Layout Plan for an Office Building - Desklib

Layout plan

Loading
Super imposed = 1.0kpa
Live Load= 3.0kpa
ρw =24+0.6 v assume v=0.5%
¿ 24 × 0.6 ×05=24.3
gk = ( 24.3× 0.2 ) +1 =5.86kpa
qk=¿ 3.0 kpa
Load combination
ULS=1.2 gk +1.5 qk
1.2 ×5.86+1.5 ×3=11.532
Slab
Panel 1 (corner slab)
2 adjacent continuous
Ly
Lx = 5000
5000 =1<2

design force per meter width, Fd:
Fd=1.2G+1.5 Q
Fd=1.2× 5.86+1.5 ×3=11.532kN /m
Minimum effective Depth for the deflection check
Lef
d ≤ K 3 K 4 [ ( ∆
Lef )Ec
Fd . ef ]1/ 3
Allowable deflection, ∆
Lef
= 1
250 Table 2.3.2
Lef =LesserofLorLn + Ds
¿ Lesserof 5000∨4700
∴ Lef =4700 mm
k cs=2−1.2 Asc
Ast
≥ 0.8 say 2.0 Cl .8 .5 .3.2
ψs =0.7 , ψ1=0.4 (Office building) From AS 1170.0:2002 Cl.4.2.2 – Table 4.1
Effective design service Load, F¿= ( 1.0+ kcs ) g+(ψs +k cs ψ1 ) q Cl.8.5.3.2
( 1.0+2.0 ) ×5.86+ ( 0.7+2.0 ×0.4 ) ×3=22.08 kN /m2
k3 =1.0 For two way slab
k 4=2.95 condition 6 Of Table 9.3.4.2
Ec=30100 case of f ' c=32 MPa Table 3.1.2

d ≥ Lef
k3 k 4
3
√ Ec ( ∆
Lef )
F¿
= 4700
1.0× 2.95
3
√ 30100 × 1
250
22.08 × 10−3
=90.5 mm
Effective depth, d= D−cover −0.5 main ̄diameter
d=200−20−0.5 ×12 where cover (A1) = 20mm, main bar=12mm
∴ dx=200−20−6=174 mm
d y=174−12=152 mm
∴ d x , d y <90.5 mm Acceptable
Moments
Ly/Lx =1.0
M*x = β 1 Fd Lx2 Cl.6.10.3.2
M*y = β 1 Fd Lx2
Coefficient βx=0.035 Table 6.10.3.2(A)
β y=0.035
Positive moment
M x
¿=0.035× 11.532× 52=10.09 kNm
M y
¿ =0.035 ×11.532 ×52=10.09 kNm
Negative moment (supports ) Cl.6.10.3.2 (B & C)

Interior support M x=1.33 × 10.09=−13.42 kNm
Exterior support M x=0.5× 10.09=−5.045 kNm
Bending design
ξ= α2 f 'c
f sy
α 2=1.0−0.003 f '
c
1.0−0.003× 32=0.904 say 0.85
γ=1.05−0.007 f '
c
1.05−0.007 ×32=0.91 say 0.85
assuming ∅=0.8
ξ= α2 f 'c
f sy
=0.85 × 32
500 =0.0544
pt =ξ− √ξ2− 2 ξM
φb d2 f sy
Bottom
In the x- direction d x=174 mm
pt =0.0544− √0.05442− 2 ×0.0544 × 10.09× 106
0.85 ×1000 ×1742 × 500 =0.00078
In the y-direction d=152mm

pt =0.0544− √ 0.05442− 2× 0.0544 ×10.09 ×106
0.85 ×1000 ×1522 ×500 =0.0001
Top
In the x-direction
pt =0.0544− √0.05442− 2 ×0.0544 × 13.42× 106
0.85 ×1000 ×1742 × 500 =0.001
In the y-direction
pt =0.0544− √0.05442− 2× 0.0544 ×5.045 ×106
0.85 ×1000 ×1522 ×500 =0.0005
ptmin=0.2 ( D
d )
2 f ' cs
f sy
ptmin=0.2 ( 200
174 )2 0.6 × √32
500 =1.793 ×10−3
pt < ptmin∴ use ptmin
In the x-direction
Ast= ptmin ×bd 1.793 ×10−3 ×1000 ×174=312 mm2 /m
∴ provide N 12 @340 mm spacing 340 mm2 /m
Checking spacing Cl.9.4.1b
Spacing¿ Lesserof [ 2 D ; 300 ] mm
¿ Lesserof [400 ; 300]

Therefore use minimum spacing of 300 mm
∴ provide N 12 @300 mm spacing 367 m m2 /m
∴Therefore the spacing is okCheck the transverse steel for cracking control
In the y-direction
Ast= ptmin ×bd 1.793 ×10−3 ×1000 ×152=273 mm2 /m
∴ provide N 12 @400 mm spacing 275m m2 / m
Checking spacing Cl.9.4.1b
Spacing ¿ Lesserof [ 2 D ; 300 ] mm
¿ Lesserof [400 ; 300]
Therefore, use minimum spacing of 300 mm
∴ provide N 12 @300 mm spacing 367 m m2 /m
∴Therefore the spacing is okCheck the transverse steel for cracking control
Location direction pt Ast N12@ s Ast actual position
Supports X 1.793 ×10−3 312 300 367 top
Mid-
region
X 1.793 ×10−3 312 300 367 bottom
supports Y 1.793 ×10−3 273 300 367 top

Mid-
region
Y 1.793 ×10−3 273 300 367 bottom
Check shear
V* = LyFd
2 = 5 ( 11.532 )
2 =28.83 KN
Vuc=β 1 β 2 β 3 bvd 0 fcv ( Ast
bvdo )
1 /3
Cl.8.2.7
β 2=β 3=1
do=200−20−6=174 mm β 1=1.1 (1.6− do
1000 )≥ 1.1
∴ β 1=1.1 (1.6− 174
1000 )=1.67 >1.1
bv=1000 mm
Ast = 367 mm2/m
f ' cv = ( f ' c )
1
3 = ( 32 )
1
3 =3.17 MPa< 4 MPa
∴Vuc=1.59 ×1.0 ×1.0 ×1000 ×154 ×3.17 ( 367
1000 ×154 ) 1
3 =103.8 kN
0.5 ∅Vuc=0.5 × 0.7× 103.8=36.33 kN ∅=0.7 Table 2.2.2
∴Vuc=¿36.33 KN¿ V∗¿ 28.83 kN
Therefore, no shear reinforcement required.
Slab
Panel 2 (edged slab)
2short edge discontinuous